Complicated Trig-quadratic

Geometry Level 1

1 − sin ⁡ 1 0 ∘ − 2 sin ⁡ 2 1 0 ∘ = sin ⁡ N ∘ 1 - \sin 10 ^ \circ - 2 \sin^2 10^\circ = \sin N^\circ , where 0 ≤ N ≤ 90 0 \leq N \leq 90 . What is N N ?


The answer is 50.

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Arron Kau by Arron Kau

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1 solution

Arron Kau Staff
May 13, 2014

Using the double angle formula, 1 − 2 sin ⁡ 2 1 0 ∘ = cos ⁡ 2 0 ∘ 1 - 2 \sin^2 10^\circ = \cos 20^\circ .

Using the sum and product formula, cos ⁡ 2 0 ∘ − sin ⁡ 1 0 ∘ = cos ⁡ 2 0 ∘ − cos ⁡ 8 0 ∘ = − 2 sin ⁡ ( 5 0 ∘ ) sin ⁡ ( − 3 0 ∘ ) = sin ⁡ 5 0 ∘ . \cos 20^\circ - \sin 10^ \circ = \cos 20^\circ - \cos 80^ \circ = - 2 \sin (50^\circ) \sin (-30^\circ ) = \sin 50^\circ. . Hence N = 50 N = 50 .

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